1 st Law of Thermodynamics

Transcription

1 Adiabatic Process

2 1 st Law of Thermodynamics total change in the internal energy of a system is the sum of the work done on the system and the heat transferred into the system. U = W on + Q in If Q = 0, then ΔU is positive if positive work is done on the system. If W = 0, then ΔU is positive if heat flows into the system.

3 Adiabatic Process No heat flows into or out of the system (Q = 0) To cause an adiabatic process, the gas can expand or be compressed very quickly or can occur in a very well insulated container.

4 When a gas expands adiabatically, the work done in the expansion comes at the expense of the internal energy of the gas, causing the temperature of the gas to drop.

5 The figure below shows P-V diagrams for these two processes.

6 The gas has a different final temperature The work done by the system is the area under the PV curve, but ends on a different isotherm

7 In adiabatic expansion, the work done by the system causes a reduction in the internal energy of the system.

8 Example One mole of ideal gas is at pressure P 0 and volume V 0. The gas then undergoes three processes: 1. The gas expands isothermally to 2V 0 while heat Q flows into the gas. 2. The gas is compressed at constant pressure back to the original volume. 3. The pressure is increased while holding the volume constant until the gas returns to its initial state.

9 Draw a P-V diagram that depicts this cycle. Label relevant points on the axes. In terms of T 0, the initial temperature, label each vertex with the temperature of the gas at that point.

10 For the remaining sections, answer in terms of T 0, Q, and R. b. Find the change in internal energy for each leg of the cycle. c. Find the work done by the gas on each leg of the cycle. d. Find the heat that flows into the gas on legs 2 and 3.

11 The two ends of the isotherm will both be at T 0. Since the product of P V is constant along an isotherm, an expansion to twice the volume implies a pressure reduction to 1/2 the original pressure. Solution

12 Applying the gas law to the lowerleft vertex then yields Solution

13 Solution (b)

14 Solution (c) 1. Since U = 0, W by = Q 2. W by = area = -1/2 P 0 V 0 =-1/2RT 0 3. W by = 0 (constant volume process)

15 Solution (d) 1. U = Q in + W on Q in = U + W by 2. Q in = -3/4RT 0 1/2RT 0 = -5/4 RT 0 3. Q in = +3/4RT 0 + 0

16 Example Heat is removed from a gas as it pressure drops from 200 kpa to 75 kpa. The gas then expands from 0.5 m 3 to 1.0 m 3. Find the work done by the gas and the heat added to the gas. isotherm 2 3

17 Solution W 1-2 = 0 W 2-3 = -P V = -75kPa x 0.5 m 3 W 2-3 = kj (done by gas) U = Q + W = 0 Q = -W = 37.5 kj of heat added to gas 2 3

18 2 nd Law of Thermodynamics Heat will not flow spontaneously from a colder body to a warmer body AND heat energy cannot be transformed completely into mechanical work. The bottom line: 1) Heat always flows from a hot body to a cold body 2) Nothing is 100% efficient

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20 Carnot Cycle A reversible isothermal gas expansion process. the ideal gas in the system absorbs an amount of heat Q in from a heat source at a high temperature T h, expands and does work on surroundings.

21 Carnot Cycle The gas continues to expand and do work on surroundings, which causes the system to cool to a lower temperature, T l

22 Carnot Cycle isothermal gas compression process. surroundings do work to the gas at T l, and causes a loss of heat, q out.

23 Carnot Cycle A reversible adiabatic gas compression process. In this process, the system is thermally insulated. Surroundings continue to do work to the gas, which causes the temperature to rise back to T h.

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25 P 1,V 1 an ideal gas draws heat from a hot reservoir at T hot and converts it to work by expanding reversibly along the T hot isotherm to P 2,V 2.

26 Then the gas is allowed to reversibly cool to T cold along an adiabat (Q = 0) to to P 3,V 3.

27 Work is then done on the gas to compress it reversibly to P 4,V 4 along the T cold isotherm, rejecting heat to the cold reservoir.

28 Finally, the gas is warmed adiabatically and reversibly back to the initial state of the gas P 1,V 1,T hot, completing the cycle.

29 Thermal Efficiency Is a measure of the ability to transform energy into work W = work, Q H = heat energy in T H = high temperature state, T C = cold temperature state

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